Proof.

Let |z|≤M.
Given ϵ>0, there is a N∈ℕ such that whenever n≥N, then
∑k=n+1∞Mk/k!<ϵ/2,
since the sum is the tail of the convergent serieseM.

Since limn→∞⁡π⁢(k,n)=1 for k, there is also a N′∈ℕ, with N′≥N, so that whenever n≥N′ and 0≤k≤N, then
|π⁢(k,n)-1|<ϵ/(2⁢eM).
(Note that k is chosen only from a finite set.)

Now, when n≥N′, we have

|∑k=0nπ⁢(k,n)⁢zkk!-∑k=0∞zkk!|

=|∑k=0n(π⁢(k,n)-1)⁢zkk!-∑k=n+1∞zkk!|

≤∑k=0n|π⁢(k,n)-1|⁢Mkk!+∑k=n+1∞Mkk!

=∑k=0N|π⁢(k,n)-1|⁢Mkk!+∑k=N+1n|π⁢(k,n)-1|⁢Mkk!+∑k=n+1∞Mkk!

<ϵ2⁢eM⁢∑k=0NMkk!+∑k=N+1nMkk!+∑k=n+1∞Mkk!

(In the middle sum, we use the bound |π⁢(k,n)-1|=1-π⁢(k,n)≤1 for all k and n.)

<ϵ2⁢eM⋅eM+ϵ2=ϵ.∎

In fact, we have proved uniform convergence of limn→∞⁡(1+zn)n
over |z|≤M.
Exploiting this fact we can also show:

(1+zn+o⁢(1n))n=(1+z+o⁢(1)n)n→∑k=0∞zkk!(pointwise, as n→∞)

Proof.

|z|<M.
Given ϵ>0, for large enough n, we have

|(1+wn)n-ew|<ϵ/2uniformly for all |w|≤M.

Since o⁢(1)→0, for large enough n we can set w=z+o⁢(1) above.
Since the exponential is continuous11follows from uniform convergence on bounded subsets of either expression for ez, for large enough n we also have |ez+o⁢(1)-ez|<ϵ/2. Thus